A gambler bought $5,000 worth of chips at a casino in denominations of $20 and $100.
A gambler bought $5,000 worth of chips at a casino in denominations of $20 and $100. That evening, the gambler lost 16 chips, and then cashed in the remainder. If the number of $20 chips lost was 2 more or 2 less than the number of $100 chips lost, what is the largest amount of money that the gambler could have received back?
Answer/Solution
$4,120
Steps/Work
In order to maximize the amount of money that the gambler kept, we should maximize # of $20 chips lost and minimize # of $100 chips lost, which means that # of $20 chips lost must be 2 more than # of $100 chips lost.
So, if # of $20 chips lost is x then # of $100 chips lost should be x-2. Now, given that total # of chips lost is 16: x+x-2=16 --> x=9: 9 $20 chips were lost and 9-2=7 $100 chips were lost.
Total worth of chips lost is 9*20+7*100=$880, so the gambler kept $5,000-$880=$4,120.
Answer: B.
So, if # of $20 chips lost is x then # of $100 chips lost should be x-2. Now, given that total # of chips lost is 16: x+x-2=16 --> x=9: 9 $20 chips were lost and 9-2=7 $100 chips were lost.
Total worth of chips lost is 9*20+7*100=$880, so the gambler kept $5,000-$880=$4,120.
Answer: B.